Work Energy Power

JEE Main level, single correct, medium

58 of 183

Question 58 of 183

An artillery piece of mass M1{M}_{1} fires a shell of mass M2{M}_{2} horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is :

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Answer: (B) M2M1\dfrac{{M}_{2}}{{M}_{1}}

As linear momentum will be conserved, ∣p⇀1∣=∣p⇀2∣\left|{p^{⇀}}_{1}\right|=\left|{p^{⇀}}_{2}\right|
Now, kinetic energy can be written as KE=p22MKE=\dfrac{{p}^{2}}{2M}.
Also pp is the same for both masses. We can write,
KE∝1mKE\propto \dfrac{1}{m}
⇒KE1KE2=p22M1p22M2=M2M1\Rightarrow \dfrac{K{E}_{1}}{K{E}_{2}}=\dfrac{\dfrac{{p}^{2}}{2{M}_{1}}}{\dfrac{{p}^{2}}{2{M}_{2}}}=\dfrac{{M}_{2}}{{M}_{1}}

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